2002 AMC 10A Problem 11

There are $3$ files of $0.8$, $12$ of $0.7$, and $30 - 15 = 15$ of $0.4$. Call the $0.8$ and $0.7$ files "large" ($15$ in all). Check what fits in $1.44$: two large files fit only if both are $0.7$ ($1.4$), and then nothing else fits; one large file leaves room for at most one $0.4$ ($0.7 + 0.4 + 0.4 = 1.5$ is too much); a disk with no large file holds at most three $0.4$ files ($1.2$).

Suppose $12$ disks were enough. Let $x, y, z$ be the numbers of disks holding two, one, or zero large files. Then $x + y + z = 12$ and $2x + y = 15$, so $x = 3 + z$ and $y = 9 - 2z$. Those disks can hold at most $y + 3z = 9 + z$ small files, and $y \ge 0$ forces $z \le 4$, so at most $13 < 15$ small files fit. Contradiction: $12$ disks fail.

Thirteen disks work: three disks each with $0.8 + 0.4$, six disks each with $0.7 + 0.7$, and four disks each with three $0.4$ files. That stores $3 + 12 + 15 = 30$ files.

The answer is $\boxed{\textbf{(B)}\ 13}$.

Packing problems are solved by a two-sided argument: a lower bound from what cannot share a disk, and a concrete arrangement that meets it. The bound must respect the integrality constraint (files do not split), so raw total size is not enough; instead, classify disks by how many large files they carry and count leftover room for the small files.

Greedy with a justification: each $0.8$ file needs its own disk, and the best use of its spare room is one $0.4$ file, so use $3$ disks for $3 \times (0.8 + 0.4)$. The remaining $12 \times 0.7 + 12 \times 0.4 = 13.2$ MB exceeds $9 \times 1.44 = 12.96$, so at least $10$ more disks are required, and $6$ disks of $0.7 + 0.7$ plus $4$ disks of three $0.4$'s achieve it: $13$ total.

• Dividing total size 16.8 MB by 1.44 to get 12 disks, ignoring that files cannot be split and that 0.8 and 0.7 cannot share a disk.
• Pairing each $0.7$ with a $0.4$ (twelve disks) and then needing four more for the $0.8$'s and leftover $0.4$'s, arriving at $15$ or $16$.